Referring to the type of an inner class in Scala

The following code tries to mimic Polymorphic Embedding of DSLs: rather than giving the behavior in Inner, it is encoded in the useInner method of its enclosing class. I added the enclosing method so that user has only to keep a reference to Inner instances, but can always get their enclosing instance. By doing this, all Inner instances from a specific Outer instance are bound to only one behavior (but it is wanted here).

abstract class Outer {
  sealed class Inner {
    def enclosing = Outer.this
  }
 def useInner(x:Inner) : Boolean
}

def toBoolean(x:Outer#Inner) : Boolean = x.enclosing.useInner(x)

It does not compile and scala 2.8 complains about:

type mismatch; found: sandbox.Outer#Inner
               required: _81.Inner where val _81:sandbox.Outer

From Programming Scala: Nested c开发者_运维问答lasses and A Tour of Scala: Inner Classes, it seems to me that the problem is that useInnerexpects as argument an Inner instance from a specific Outer instance.

What is the true explanation and how to solve this problem ?

I suppose the type Inner is like the type this.Inner. Outer#Inner is independent of the outer instance (not a path-dependent type).

abstract class Outer {
  sealed class Inner {
    def enclosing = Outer.this
  }
  def useInner(x:Outer#Inner) : Boolean
}

def toBoolean(x:Outer#Inner) : Boolean = x.enclosing.useInner(x)

The problem is as you describe, that useInner is expecting an Inner of a specific Outer instance. Since enclosing returns a generic Outer, there is really no way to tie both together that I know of.

You can force it, however:

def toBoolean(x: Outer#Inner): Boolean = {
  val outer = x.enclosing
  outer.useInner(x.asInstanceOf[outer.Inner])
}

You can also define your member like this:

def useInner(x:Outer#Inner) : Boolean

Or you can write like this:

abstract class Outer {
    class InnerImpl {
        def enclosing = Outer.this
    }
    final type Inner = Outer#InnerImpl
    def useInner(x:Inner) : Boolean
}